0,3

Absolute value of the alternating sum of Catalan Numbers. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 03 2006

Sums of two consecutive terms are A032357[n-1]+A032357[n]=1,2,5,14,42..=A000108[n] Catalan Numbers. The prime p divides a((p-3)/2) for p=11,19,29,31,41,59,61,71..=A045468[n] Primes congruent to {1, 4} mod 5. Prime p divides a(2p+1) for p=5,11,19,29,31,41,59,61,71..=A038872[n] Primes congruent to {0, 1, 4} mod 5. Also odd primes where 5 is a square mod p. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 03 2006

Hankel transform is F(2n+1). - Paul Barry (pbarry(AT)wit.ie), Jul 22 2008

Equals INVERTi transform of A000958 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Apr 10 2009]

Inverse binomial transform of A002212 . [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Sep 17 2009]

G.f.: c(x)/(1+x), where c(x) = g.f. for Catalan numbers.

Sum((-1)^(n-k)*C(k), k=0..n), C(k)=A00108(k), Catalan numbers

a(n)=((-1)^(n+1)-binomial(2*(n+1), n+1)*sum((-5)^k*binomial(n+1, k)/binomial(2*k, k), k=0..n+1))/2

Also: a(n)=C[2n, n]/(n+1)-a(n-1)=A000108[n]-a(n-1) with a(0)=1. - Labos E. (labos(AT)ana.hu), Apr 26 2003

a(n) = Sum[ (-1)^(k+n)*CatalanNumber[k], {k,0,n}]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 03 2006

Table[Sum[(-1)^(k+n)*CatalanNumber[k], {k, 0, n}], {n, 0, 60}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 03 2006

Cf. A000108.

Cf. A000108, A014137, A014138, A033297, A045468, A038872, A064739.

A000958 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Apr 10 2009]

Adjacent sequences: A032354 A032355 A032356 this_sequence A032358 A032359 A032360

Sequence in context: A038987 A142957 A080155 this_sequence A144056 A062630 A159458

easy,nonn

Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de)

More terms from Christian G. Bower (bowerc(AT)usa.net), Apr 15 1998.

More terms from Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 03 2006